The upper chamber of an hour-glass is a cone of radius 3 inches and height 10 inches and, if full, it requires exactly one hour to empty. Assuming that the sand falls through the aperture at a constant rate, how fast is the level falling when:
a) the depth of the sand is 6 inches?
b) 105/2 minutes have elapsed from the time when the hour-glass was full in the upper chamber?
v = π/3 * 3^2 * 10 = 30π
so, dv/dt = -30π in^3/hr = -π/2 in^3/min
When the sand has a depth of y inches, the surface has a radius of r = (3/10) y
So, at that point,
v = π/3 y^2 r = π/10 y^3
dv/dt = 3π/10 y^2 dy/dt
-π/2 = 3π/2 * 6^2 dy/dt
dy/dt = -1/108 in/min
For the 2nd part, find r, using
v(t) = 30π - π/2 t
then use it as in the first part.
Why you have v(t) = 30π - π/2 t???
I have that formula because, as I showed at the very first, the volume starts at 30π ion^3, and drops at a rate of π/2 in^3/min.
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I swapped y and r in the volume formula. So, doing it right,
v = π/3 r^2 y = 3π/100 y^3
dv/dt = 9π/100 y^2 dy/dt
-π/2 = 9π/100 * 6^2 dy/dt
dy/dt = -25/162 in/min
Or, -250/27 = -9.26 in/hr
Since it is negative, the level is falling at 9.26 in/hr
To find the rate at which the level of sand in the upper chamber of the hourglass is falling, we can use related rates. Let's go through each part.
a) When the depth of the sand is 6 inches:
To determine how fast the level is falling at this specific depth, we need to find the rate of change of the depth with respect to time.
Given:
Radius of the cone (upper chamber) = 3 inches
Height of the cone (upper chamber) = 10 inches
First, let's express the volume of the sand in terms of the depth.
The volume of a cone can be expressed as V = (1/3) * π * r^2 * h, where r is the radius and h is the height.
In our case, since the upper chamber is a cone, the depth of the sand will also be the height of the cone.
Let's express the depth as 'x'. Therefore, at any given time, the depth of the sand will be 'x' inches, and we need to find the rate at which x is changing.
Now, we can express the volume of the sand in terms of 'x':
V = (1/3) * π * (3^2) * x
Since the hourglass takes exactly one hour to empty, the volume of sand flowing out per unit of time is constant. Therefore, the rate of change of volume will be constant.
We can express this rate as dV/dt, where dV is the change in volume and dt is the change in time.
Now, let's differentiate the volume equation with respect to time to find dV/dt:
dV/dt = (1/3) * π * (3^2) * dx/dt
We know that dx/dt represents the rate at which the depth of the sand is changing over time. This is the value we're trying to find.
At the depth of 6 inches, or x = 6, we need to find dx/dt.
To find dx/dt, we can rearrange the equation:
dx/dt = (3 * dV/dt) / (π * (3^2))
Since we know the rate at which the volume is changing is constant for the hourglass, we can denote it as dV/dt = k, where k is a constant.
Plugging in the values, we have:
dx/dt = (3 * k) / (π * (3^2))
Therefore, the rate at which the level of sand is falling when the depth is 6 inches is given by:
dx/dt = (3 * k) / (π * 9)
b) When 105/2 minutes have elapsed:
We'll use similar calculations as in part a) but substitute the time value.
Given that 1 hour = 60 minutes, we have the elapsed time as 105/2 minutes. This is the value of t that we need to use.
Repeating the steps from part a), we'll find that:
dx/dt = (3 * k) / (π * 9)
Therefore, the rate at which the level of sand is falling when 105/2 minutes have elapsed is given by:
dx/dt = (3 * k) / (π * 9)
Note: The constant 'k' represents the rate of sand flowing through the aperture consistently and can be determined if we know the specifics of the hourglass.